A random variable $X$ has Poisson distribution with mean 2. Then $P(X>1.5)$ equals

A random variable $X$ has Poisson distribution with mean 2. Then $P(X>1.5)$ equals
  1. $\frac{2}{\mathrm{e}^2}$
  2. 0
  3. $1-\frac{3}{\mathrm{e}^2}$
  4. $\frac{3}{\mathrm{e}^2}$

Solution

$ \begin{aligned} & P(x=k)=e^{-\lambda} \frac{\lambda^k}{k !} \\ & P(x \geq 2)=1-P(x=0)-P(x=1) \\ & =1-e^{-\lambda}-e^{-\lambda}\left(\frac{\lambda}{1 !}\right) \\ & =1-\frac{3}{e^2} . \end{aligned} $

Asked in: JEE Main 2005

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